MAT 167: Advanced Linear Algebra

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Lynette Haynes
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1 < Proem 1 (15 pts) MAT 167: Advanced Linear Agera Fina Exam Soutions (a) (5 pts) State the definition of a unitary matrix and expain the difference etween an orthogona matrix and an unitary matrix. Soution: A unitary matrix is a square matrix of size whose coumn vectors form an orthonorma asis for. In other words, a matrix is unitary if. In other words,. An orthogona matrix is its counterpart for. In other words, an orthogona matrix is a square matrix of size whose coumn vectors form an orthonorma asis for and satisfies!. () (5 pts) Prove if is unitary, then for any "#, we have % "&%()%"&*. Soution: e simpy use the definitions of the 2norm for vectors, the innerproduct, and a unitary matrix. % "! +, " "/./12 "4 2 "/ " "5" "6 +2""/./7%"& Since oth % "&% and %"&* are nonnegative, we can take nonnegative square roots of oth sizes of the aove equaity to get % ":;4)%"&*. (c) (5 pts) Show that the coumn vectors of any unitary matrix 9 are ineary independent. Soution: Let us consider a inear comination of coumns of : < CD is the G th coumn vector of. Using the notation, HIJ < where FE can e rewritten as: HC Mutipying from eft on the oth sides yieds: 9K < L, the aove Therefore, K9 HMHMC are ineary independent.

2 Œ Œ Score of this page: 2 Proem 2 (15 pts) Consider the foowing matrix: SR;TVUX ZU\[^]_ O&P Ù [a]_ R;TVUXc _dfeghid (a) ( pts) Show that O:P is an orthogona matrix. Soution: This can e easiy done y computing O!PO P () ( pts) Determine the Fourier expansion of " are the coumn vectors of O:P. Soution: Let the Fourier expansion of " e Hqwhere < O P "6 < R;TVUX U\[^]_ < O P O&P. k with respect to the asis whose vectors mu\[a]n RKToUp and < are the Fourier coefficients of ", and Hr s < O&P < \t. Since these coumn vectors form an orthonorma asis, the Fourier coefficients can e computed y mutipying on oth sides from eft: O P "5 Thus, the Fourier expansion in this case is: (c) ( pts) Show that O:P *( Soution: The 2norm of any matrix u R;TVUp RKToUp mu\[a]n R;TVUp Ù [a]_ O P O&P 5U\[a]n regardess of the vaues of. HMH ZU\[^]n R;TVUp is simpy the argest singuar vaue v need to do is to compute the argest eigenvaue of u&xu. In this case, O P O&P k {z wqx%y {z {z 1 z Thus the eigenvaues are: O&P we concude: * O&P (d) ( pts) Compute.. Thus the argest singuar vaue v independent of the vaues of. } of u. Thus a we, thus. Therefore Soution: Reca the 1norm of a matrix u is the maximum of the 1norm of the coumn vectors of u, i.e., *u~ ƒ E ˆ *um> Š G. Thus, in this case, the 1norm of the 1st coumn vector and that of the 2nd coumn vector are the same, i.e., O&P R;TVUp 7Œ U\[^] Œ

3 œ e e Œ e Œ Œ Œ O&P R;TVU U\[^]_ š7 )Œ Œ d First, notice that V R;TVU d ZÙ [a] using the asic trig. identities, such as V Ù [a] d RKToU, V. This means that d is periodic with period. Therefore, we R;TVUpœ ony need to Ù check [a]_œ this function over the interva e fe7d. In this interva, ceary and. Thus, R;TVUp U\[^]_ R;TVU fd Thus it is cear that d. The equaity hods if in this d interva. ow, it is aso cear that in this interva, and the equaity hods if. Using the periodicity of e, we concude that nd nd Ÿd and _d the first d equaity d hods if F, and the second equaity hods if. Score of this page: (e) ( pts) Show that O:P O&P %Ž7 for any vaue of. Soution: Reca the norm of a matrix u is the maximum of the 1norm of the row vectors of u, i.e., *u~%ž 1~ ƒ ˆ a *um \K B. Thus, in this case, the 1norm of the 1st row vector and that of the 2nd row vector are the same, i.e., O&P R;TVUp %Ž)Œ (f) Bonus proem (5 pts) Show that e hod O:P U\[a] O&P Œo7. For what vaues of do the equaities [Hint: R;TVUp Check Ù [a]n the periodicity R;TVU :6d of the 1norm in this case, and then use the trig. identity:.] Soution: Let }

4 L ( + DB. ( J F L () o (1J L O Score of this page: 4 Proem (1 pts) Compute the R factorization of the foowing matrix: u Soution: Use the GramSchmidt procedure here. Let Step 1: ª Step 2: ª Step : ª Finay: e get: 1J L 7; J * DB* e the coumn vectors of u. L J ˆ ( + 9./ ( + 9./ ( () o 1J L J L

5 A Score of this page: 5 Proem 4 (1 pts) For every innerproduct space ± over the scaar fied, show that if ² ³ ±, then ² µ is a suspace of ± even if ² itsef is not a suspace of ±. Soution: Take any " ² µ and any ¹. Take any memer º ². Then, y definition of» space we must have +," º}._ +2 4\º}.. ow, the iinearity of the innerproduct, we have: Thus " derived independenty from whether ² +," / º}. +2" º. +2 ( º./ +2¹¼" º./ ¹(+," º./ ¹ {² µ and ¹½"} {²#µ. This impies that ² µ is a suspace of ±, and this was is a suspace of ± or not. This competes the proof.

6 Score of this page: 6 Proem 5 (1 pts) Consider a inear system u&"#1¾, where u1 that satisfies u C. /, " Then show that this system of equations is consistent if and ony if +k¾ À. [Hint: Consider the orthogona decomposition: ÁMSu!DÂiÃÄ u:.], and ¾. for any Soution: 2@Å : Suppose u:"# ¾ is a consistent system. This means that ¾Ä 6ÁMSu!. ow, any vector # with u 6}C means that # rãæsu. But we know that ÁMSu!4» ÃÆSu in. Thus, ceary, ¾», i.e., +k¾q\ À./. 2Ç : Suppose ¾Z»È for any gãæsu!. Then, either ¾ZC or ¾6 gá u:. If ¾r C, then u&" C aways has a soution ")C. If ¾r Á u:, then this means that u&" ¾ is consistent. Thus, either way, u&" }¾ is consistent.

7 Ò s s z Score of this page: 7 Proem 6 (1 pts) Consider the foowing matrix: ué (a) (5 pts) Compute the Singuar Vaue Decomposition (SVD) of u. Soution: Foows the usua strategy. Thus, we have z wqx;y u u Êz {z Êz {z Êz {z z 2 Êz {z {z K` ;, Êz z~ê. Therefore, the singuar vaues are: V z ow, to determine Ë z, we sove the eigenvaueeigenvector proems. For k Thus we can set s Thus z we have For : Thus we can set s ength, z it ecomes For : Thus, we can set s t Ïs } Ð s t{ s ` t ( s t. ow, to compute Ñ, we use the formua: Ò E t, ut we need to normaize it to have a unit ength. t.. t, and after the normaization, to have a unit t. t, and again after the normaization, it ecomes ÓkÔ u E for v EMÕ Ð Ð Ð. Thus, :

8 v Ò A t Score of this page: Thus the SVD of u is: Ò4 Ð Ð Ð () (5 pts) Compute the rank approximation of u. Soution: The rank approximation of u is of course s Ð s Ð Ð Ð t

9 s z Score of this page: 9 Proem 7 (15 pts) Consider the matrices u Ö1 (a) (5 pts) hat is the orthogona proector Soution: t under ØcÉuM u u: u Therefore the image of s Ù () (1 pts) Same questions for Ö. onto Á u:, and what is the proection of the vector t under is: [Hint: Compute ÖMÚ, i.e., the pseudoinverse of Ö, and construct an orthogona proector using Ö Ú.] Soution: In this case, Ö is not of fu rank and Ö Ö Thus, we cannot do ÀÛ ÜÖÄ ÖÝÖ >Ö. Instead we need to compute the pseudoinverse ÖMÚ and construct the orthogona proector Û#7ÖMÖMÚ. To do so, we need to compute first the SVD of Ö. Êz wqx%y {z {z z So, and v, v z. As for the eigenvector for, k So, Þs t, and after the z normaization, we get: Ïs ow for the eigenvector for, t.

10 Ò s Score of this page: 1 Thus, with the normaization in mind, we get ( t. For Ñ, we use the formua Ò ÓKß u to get: k e need to find Ò4 and Ò( orthogona to Ò with *Ò4V~à*Ò4 o~. So, y ooking the components of Ò, we can easiy find, for exampe, Òm s t. Then, Ò can e Ò4 Ïs t. Thus, the SVD of Ö is: Ö Therefore, Ö Ú ËMá Ú Ñ > > k Therefore, Therefore the image of s ¼ÛmÖMÖ Ú Ù t under is:

11 ¹ â ã ã ¹ F â s ä Ð ä â Score of this page: 11 Proem (15 pts) Suppose we are given three data points in the sense of the east squares. (a) (5 pts) rite a system of equation in the form u:" ¾, where "5Ùs ¹ passes through a these three points. Soution: Since U\[^] e easiy written as: when _d d U\[^] * F d _d, F % ow, we want to find the est function to fit these points in the form of 1¹ #â in t as if this function, respectivey, the system of equation can () (5 pts) Sove the eastu\[^] squares proem using the norma equation, and write the soution in the. form of ¹ {â Soution: This is simpy can e soved y Su u: u ¾ Thus the east square soution is:! Ù [a] > o (c) (5 pts) Sove the east squares proem using the reduced R factorization and confirm that your hand computed resut agrees with that of (). Soution: Let the coumn vectors of u e this case is: ã ; _ and. Then the reduced R factorization of u ã s t in. ä + 9./ ã _ ã ( t Thus, u O æå å

12 ¹ â O å ¾ å O å Score of this page: 12 ow, we transform the inconsistent system u:"i ¾ using the orthogona proector å å a consistent system: u:"6 æå :D¾. Using the R factorization of u, this simpifies to: "5 M¾ å : to Therefore, "6 M¾ å Ð o Therefore, the east squares soution ased on the reduced R factorization is the same as that y the norma equation. (d) Bonus Proem (5 pts) After fitting the east squares soution you computed in () and (c) to the actua data, compute the residua (or error) vector and its ength in 2norm. Soution: The error vector is simpy, u:" o o o o Thus its 2norm is: o o Ð

Matrices and Determinants

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